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Let $V$ be the vector space of all $3\times 3$ matrices.

Let $W=\{A\in V \mid A^{-1} \text{exists}\}$

Is $W$ a vector space?

I'm having a hard time solving this problem. I know we must apply the 10 axioms, but I am not sure how for this particular problem.
Thanks.

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    What is the zero? – 2017-02-23
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    if W is a vector space then it must be closed under addition. It is possible to add two non-singular matrices to get a singular matrix? – 2017-02-23
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    To be precise, you should state what your addition and multiplication operations, on $W$, are. – 2017-02-23
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    @DougM I can't think of one, but yes... I believe it is possible. – 2017-02-23
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    have you done anything with eigenvalues yet? $\det (A - \lambda I) = 0$ is a very important formula there. $A$ is frequently non-singular and $\lambda I$ is definitely non-singular – 2017-02-23
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    We've just barely touched base with eigenvalues. JohnD did a good job of simplifying it for me. Thank you all. – 2017-02-23

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Since you didn't mention how the required operations (addition and scalar multiplication) are defined on $W$, I assume you mean they are the standard operations. You also need to specify what your set of scalars is.

For $W$ to be a subspace of $V$, the Subspace Theorem requires (1) the existence of a zero vector in $W$, (2) $W$ is closed under addition, and (3) $W$ is closed under scalar multiplication.

Any of these is easy to violate. The zero vector would be the $3\times 3$ matrix of zeros, which is not invertible, so it does not live in $W$ and thus (1) fails.

You could also show (2) fails by considering the $3\times 3$ identity matrix, $I$. $I\in W$ and $-I\in W$ but $I+(-I)=0$ (the zero matrix) and thus is not invertible, so it cannot be in $W$. Thus, (2) fails.

You could show (3) fails by just picking any matrix $M\in W$ and multiplying it by the zero scalar to obtain the zero matrix, which we've established is not in $W$. Thus, (3) fails.